Three-Phase System Three phase systems Faculty of IIITN Ratnakar N

1.1: Review of Single-Phase System The Sinusoidal voltage v1(t) = Vm sin wt i v1 v2 Load AC generator

1.1: Review of Single-Phase System The Sinusoidal voltage v(t) = Vm sin wt where Vm = the amplitude of the sinusoid w = the angular frequency in radian/s t = time Three phase systems Faculty of IIITN Ratnakar N

The angular frequency in radians per second

A more general expression for the sinusoid (as shown in the figure): v2(t) = Vm sin (wt + q) where q is the phase

A sinusoid can be expressed in either sine or cosine form A sinusoid can be expressed in either sine or cosine form. When comparing two sinusoids, it is expedient to express both as either sine or cosine with positive amplitudes. We can transform a sinusoid from sine to cosine form or vice versa using this relationship: sin (ωt ± 180o) = - sin ωt cos (ωt ± 180o) = - cos ωt sin (ωt ± 90o) = ± cos ωt cos (ωt ± 90o) = + sin ωt

Sinusoids are easily expressed in terms of phasors Sinusoids are easily expressed in terms of phasors. A phasor is a complex number that represents the amplitude and phase of a sinusoid. v(t) = Vm cos (ωt + θ) Time domain Phasor domain Time domain Phasor domain

Time domain Phasor domain or θ or v2(t) = Vm sin (wt + q)

1.1.1: Instantaneous and Average Power The instantaneous power is the power at any instant of time. p(t) = v(t) i(t) Where v(t) = Vm cos (wt + qv) i(t) = Im cos (wt + qi) Using the trigonometric identity, gives

The average power is the average of the instantaneous power over one period.

The effective value is the root mean square (rms) of the periodic signal. The average power in terms of the rms values is Where

1.1.2: Apparent Power, Reactive Power and Power Factor The apparent power is the product of the rms values of voltage and current. The reactive power is a measure of the energy exchange between the source and the load reactive part.

The power factor is the cosine of the phase difference between voltage and current. The complex power:

1.2: Three-Phase System 3-phase system will produce output of 1.5times of 1phase motor output. The increase in efficiency from 1phase to 2phase is 41.4% The increase in efficiency from 1phase to 3phase is 50% A three-phase system is superior economically and advantage, and for an operating of view, to a single-phase system. In a balanced three phase system the power delivered to the load is constant at all times, whereas in a single-phase system the power pulsates with time.

eB = EmB sin (wt -240o) = EmBsin (wt +120o) The instantaneous e.m.f. generated in phase R, Y and B: eR = EmR sin wt eY = EmY sin (wt -120o) eB = EmB sin (wt -240o) = EmBsin (wt +120o) 15

IR VR ZR ER IN EB VB VY EY ZB ZY IY IB Three-phase AC generator Three-phase Load

eR = EmR sin ωt Phase voltage The instantaneous e.m.f. generated in phase R, Y and B: eR = EmR sin ωt eY = EmY sin (ωt -120o) eB = EmB sin (ωt -240o) = EmBsin (ωt +120o) In phasor domain: 120o ER = ERrms 0o 0o EY = EYrms -120o -120o EB = EBrms 120o Magnitude of phase voltage ERrms = EYrms = EBrms = Ep

Line voltage IR ERY VR ZR ER IN EB VB VY EY ZB ZY IY IB ERY = ER - EY Three-phase AC generator Line voltage IR ERY VR ZR ER IN EB VB VY EY ZB ZY IY IB ERY = ER - EY Three-phase Load

Line voltage = Ep 0o - Ep -120o = 1.732Ep ERY 30o = √3 Ep = EL ERY -EY 120o 0o -120o ERY = ER - EY

Line voltage IR VR ZR ER IN EB VB VY EY ZB ZY IY IB EYB EYB = EY - EB Three-phase AC generator Line voltage IR VR ZR ER IN EB VB VY EY ZB ZY IY IB EYB EYB = EY - EB Three-phase Load

Line voltage = Ep -120o - Ep 120o = 1.732Ep EYB -90o = √3 Ep = EL 120o 0o -120o -EB EYB EYB = EY - EB

Line voltage IR VR ZR ER EBR IN EB VB VY EY ZB ZY IY IB EBR = EB - ER Three-phase AC generator Line voltage IR VR ZR ER EBR IN EB VB VY EY ZB ZY IY IB EBR = EB - ER Three-phase Load

For star connected supply, EL= √3 Ep Line voltage = Ep 120o - Ep 0o = 1.732Ep EBR 150o EBR 120o 0o 150o = √3 Ep = EL -ER -120o For star connected supply, EL= √3 Ep EBR = EB - ER 24

Phase voltages It can be seen that the phase voltage ER is reference. ER = Ep 0o EY = Ep -120o EB = Ep 120o 120o 0o Line voltages -120o ERY = EL 30o EYB = EL -90o EBR = EL 150o

Or we can take the line voltage ERY as reference. Phase voltages ER = Ep -30o EY = Ep -150o EB = Ep 90o Line voltages ERY = EL 0o EYB = EL -120o EBR = EL 120o Or we can take the line voltage ERY as reference. 26

Delta connected Three-Phase supply Three-phase AC generator Delta connected Three-Phase supply IR ERY VR ZR ER EB VB VY EY ZB ZY IY IB Three-phase Load ERY = ER = Ep 0o

Delta connected Three-Phase supply Three-phase AC generator Delta connected Three-Phase supply IR VR ZR ER EB VB EBR VY EY ZB ZY IY IB EYB Three-phase Load For delta connected supply, EL= Ep

Connection in Three Phase System 4-wire system (neutral line with impedance) 3-wire system (no neutral line ) 4-wire system (neutral line without impedance) 3-wire system (no neutral line ), delta connected load Star-Connected Balanced Loads a) 4-wire system b) 3-wire system Delta-Connected Balanced Loads a) 3-wire system

4-wire system (neutral line with impedance) Three-phase AC generator IR VR ZR ER ZN IN EB VB VN VY EY ZB ZY IY IB 1.1 Three-phase Load Voltage drop across neutral impedance: VN = INZN

4-wire system (neutral line with impedance) Three-phase AC generator Applying KCL at star point IR VR ZR ER ZN IN EB VB VN VY EY ZB ZY IY IB 1.2 Three-phase Load IR + IY + IB= IN

4-wire system (neutral line with impedance) Three-phase AC generator Applying KVL on R-phase loop IR VR ZR ER ZN IN EB VB VN VY EY ZB ZY IY IB Three-phase Load

4-wire system (neutral line with impedance) Three-phase AC generator Applying KVL on R-phase loop IR VR ZR ER ZN IN VN ER – VR – VN = 0 ER – IRZR – VN = 0 Thus ER – VN 1.3 IR = Three-phase Load ZR

4-wire system (neutral line with impedance) Three-phase AC generator Applying KVL on Y-phase loop IR VR ZR ER ZN IN EB VB VN VY EY ZB ZY IY IB Three-phase Load

4-wire system (neutral line with impedance) Three-phase AC generator Applying KVL on Y-phase loop EY – VY – VN = 0 Thus EY – VN 1.4 IY = EY – IYZY – VN = 0 ZY ZN IN VN VY EY ZY IY Three-phase Load

4-wire system (neutral line with impedance) Three-phase AC generator Applying KVL on B-phase loop EB – VB – VN = 0 Thus EB – VN 1.5 IB = EB – IBZB – VN = 0 ZB ZN IN EB VB VN ZB IB Three-phase Load

4-wire system (neutral line with impedance) Substitute Eq. 1.2, Eq.1.3, Eq. 1.4 and Eq. 1.5 into Eq. 1.1: IR + IY + IB= IN ER – VN EY – VN EB – VN VN = + + ZR ZY ZB ZN ER – VN EY – VN + EB – VN = VN ZN ZR ZY ZB ER ZR + EY ZY EB ZB = 1 ZN VN

4-wire system (neutral line with impedance) = ER ZR + EY ZY EB ZB 1 ZN 1.6 VN ER ZR + EY ZY EB ZB = 1 ZN VN

4-wire system (neutral line with impedance) VN is the voltage drop across neutral line impedance or the potential different between load star point and supply star point of three-phase system. = ER ZR + EY ZY EB ZB 1 ZN 1.6 VN We have to determine the value of VN in order to find the values of currents and voltages of star connected loads of three-phase system.

Example IR EL = 415 volt VR ZR = 5 Ω ER ZN =10 Ω IN EB VB VN ZY= 2 Ω EY ZB = 10 Ω IY IB Find the line currents IR ,IY and IB. Also find the neutral current IN. Three-phase Load

3-wire system (no neutral line ) Three-phase AC generator IR VR ZR ER ZN IN EB VB VN VY EY ZB ZY IY IB Three-phase Load

3-wire system (no neutral line ) Three-phase AC generator IR VR ZR ER VN EB VB VY EY ZB ZY IY IB Three-phase Load No neutral line = open circuit , ZN = ∞

3-wire system (no neutral line ) ER ZR EY ZY EB ZB + + 1.6 VN = 1 1 ZR 1 ZY 1 ZB + + + ZN ZN = ∞ ∞ 1 ∞ =

3-wire system (no neutral line ) ER ZR EY ZY EB ZB + + 1.7 VN = 1 ZR + ZY ZB

Example IR EL = 415 volt VR ZR = 5 Ω ER EB VB VN ZY= 2 Ω EY ZB = 10 Ω IY IB Find the line currents IR ,IY and IB . Also find the voltages VR, VY and VB. Three-phase Load

3-wire system (no neutral line ),delta connected load Three-phase AC generator IR VR ZR ER EB VB VY EY ZB ZY IY IB Three-phase Load

3-wire system (no neutral line ),delta connected load Three-phase AC generator IR Ir ER VRY VBR ZRY ZBR EB Ib ZYB EY Iy IY VYB IB Three-phase Load

3-wire system (no neutral line ),delta connected load Three-phase AC generator IR Ir ERY =VRY ER VRY VBR ZRY EBR ZBR =VBR EB EY Ib ZYB Iy IY VYB IB EYB =VYB Three-phase Load

3-wire system (no neutral line ),delta connected load Phase currents VRY ZRY ERY ZRY EL ZRY 30o Ir = = = VYB ZYB EYB ZYB EL ZYB -90o Iy = = = VBR ZBR EBR ZBR EL ZBR 150o Ib = = =

3-wire system (no neutral line ),delta connected load Three-phase AC generator IR Ir Line currents ERY =VRY ER IR = Ir - Ib VRY VBR ZRY EL ZRY 30o EBR EL 150o ZBR =VBR - EB = EY ZBR Ib ZYB Iy IY IY = Iy - Ir VYB EL EL ZYB -90o 30o EYB =VYB IB - = ZRY Three-phase Load

3-wire system (no neutral line ),delta connected load Three-phase AC generator IR Ir Line currents ERY =VRY ER IB = Ib - Iy VRY VBR ZRY EL ZBR 150o EBR EL -90o ZBR =VBR - EB = EY ZYB Ib ZYB Iy IY VYB EYB =VYB IB Three-phase Load

Star to delta conversion ZRY = ZRZY + ZYZB ZBZR ZB ZYB ZR ZBR ZY ZR ZBR ZRY ZY ZB ZYB

Use star-delta conversion. Example Use star-delta conversion. IR EL = 415 volt VR ZR = 5 Ω ER EB VB VN ZY= 2 Ω EY ZB = 10 Ω IY IB Find the line currents IR ,IY and IB . Three-phase Load

4-wire system (neutral line without impedance) Three-phase AC generator IR VR ZR ER ZN 0 Ω IN = EB VB VN VY EY ZB ZY IR IB Three-phase Load VN = INZN = IN(0) = 0 volt

4-wire system (neutral line without impedance) For 4-wire three-phase system, VN is equal to 0, therefore Eq. 1.3, Eq. 1.4, and Eq. 1.5 become, ER ER – VN 1.3 IR = ZR EY – VN EY 1.4 IY = ZY EB EB – VN 1.5 IB = ZB

Example IR EL = 415 volt VR ZR = 5 Ω ER IN EB VB VN ZY= 2 Ω EY ZB = 10 Ω IY IB Find the line currents IR ,IY and IB . Also find the neutral current IN. Three-phase Load

eB = EmB sin (wt -240o) = EmBsin (wt +120o) The instantaneous e.m.f. generated in phase R, Y and B: eR = EmR sin wt eY = EmY sin (wt -120o) eB = EmB sin (wt -240o) = EmBsin (wt +120o)

1.4: Phase sequences RYB and RBY (a) RYB or positive sequence VR leads VY, which in turn leads VB. This sequence is produced when the rotor rotates in the counterclockwise direction.

(b) RBY or negative sequence VR leads VB, which in turn leads VY. This sequence is produced when the rotor rotates in the clockwise direction.

1.5: Connection in Three Phase System 1.5.1: Star Connection a) Three wire system

Star Connection b) Four wire system

Wye connection of Load

1.5.2: Delta Connection

Delta connection of load

1.6: Balanced Load Connection in 3-Phase System

Wye-Connected Balanced Loads b) Three wire system Example IR EL = 415 volt VR ZR = 20 Ω ER EB VB ZY= 20 Ω VN EY ZB = 20 Ω IY IB Find the line currents IR ,IY and IB . Also find the voltages VR, VY and VB. Three-phase Load

Wye-Connected Balanced Loads b) Three wire system VN = = 0 volt VR = ER VY = EY VB = EB

1.6.1: Wye-Connected Balanced Loads a) Four wire system Example IR EL = 415 volt VR ZR = 20 Ω ER IN EB ZY= 20 Ω VB VN EY ZB = 20 Ω IY IB Find the line currents IR ,IY and IB . Also find the neutral current IN. Three-phase Load

1.6.1: Wye-Connected Balanced Loads a) Four wire system For balanced load system, IN = 0 and Z1 = Z2 = Z3

Wye-Connected Balanced Loads b) Three wire system

1.6.2: Delta-Connected Balanced Loads Phase currents: Line currents:

1.7: Unbalanced Loads

1.7.1: Wye-Connected Unbalanced Loads Four wire system For unbalanced load system, IN  0 and Z1  Z2  Z3

1.7.2: Delta-Connected Unbalanced Loads Phase currents: Line currents:

Power in a Three Phase System

The three phase power is equal the sum of the phase powers Power Calculation The three phase power is equal the sum of the phase powers P = PR + PY + PB If the load is balanced: P = 3 Pphase = 3 Vphase Iphase cos θ

1.8.1: Wye connection system: I phase = I L and Real Power, P = 3 Vphase Iphase cos θ Reactive power, Q = 3 Vphase Iphase sin θ Apparent power, S = 3 Vphase Iphase or S = P + jQ

1.8.2: Delta connection system VLL= Vphase P = 3 Vphase Iphase cos θ

The Single-Phase Equivalent By now it should be apparent that if you know the solution for one phase of a balanced system, you effectively know the solution for all three phases.

Basic Three-Phase Relationships

Tutorial Problems

Practice problem 1

Three phase systems Faculty of IIITN Ratnakar N

Practice problem 1 Three phase systems Faculty of IIITN Ratnakar N

Three phase power measurement

Power measurement In a four-wire system (3 phases and a neutral) the real power is measured using three single-phase watt-meters.

Three Phase Circuit Four wire system, Each phase measured separately

watt-meter connection Current coil (low impedance) voltage coil (high impedance) 96

a) Four wire system Example WR IR VR ER EL = 415 volt IN EB VB EY VN ZR = 5 30o Ω ER EL = 415 volt IN EB VB ZY = 10 90o EY VN Ω WY ZB = 20 45o Ω IY IB WB Three-phase Load Find the three-phase total power, PT.

b) Three wire system Example WR IR VR ER EL = 415 volt EB EY VN WY IY ZR = 5 30o Ω ER EL = 415 volt EB ZY = 10 90o EY VN Ω WY ZB = 20 45o Ω IY IB WB Three-phase Load Find the three-phase total power, PT.

b) Three wire system Example WR IR VR ER EL = 415 volt EB VB EY VN WY ZR = 5 30o Ω ER EL = 415 volt EB VB ZY = 10 90o EY VN Ω WY ZB = 20 45o Ω IY IB WB Three-phase Load Find the three-phase total power, PT.

Three Phase Circuit Three wire system, The three phase power is the sum of the two watt-meters reading

Proving: The three phase power (3-wire system) is the sum of the two watt-meters reading Instantaneous power: pA = vA iA pB = vB iB pC = vC iC pT = pA + pB + pC = vA iA + vB iB +vC iC = vA iA + vB iB +vC iC = vA iA + vB (-iA -iC) +vCiC 101

Proving: The three phase power (3-wire system) is the sum of the two watt-meters reading Instantaneous power: pT = vA iA + vB (-iA –iC) +vCiC = (vA – vB )iA + (vC – vB )iC = vAB iA + vCBiC pT = pAB + pCB 102

Phasor diagram for a three-phase balanced star-connected circuit

Comments on Two Wattmeter Readings(V.V.IMP)

Power measurement In a four-wire system (3 phases and a neutral) the real power is measured using three single-phase watt-meters. In a three-wire system (three phases without neutral) the power is measured using only two single phase watt-meters. The watt-meters are supplied by the line current and the line-to-line voltage. 107

Q1 Q2

Practice Problems P2 P1

P3 P4

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